Problem: Solve for $x$ and $y$ using elimination. $\begin{align*}2x-8y &= -8 \\ 2x+4y &= -5\end{align*}$
Solution: We can eliminate $x$ when its corresponding coefficients are negative inverses. Recalling our knowledge of least common multiples, multiply the top equation by $-1$ and the bottom equation by $1$ $\begin{align*}-2x+8y &= 8\\ 2x+4y &= -5\end{align*}$ Add the top and bottom equations. $12y = 3$ Divide both sides by $12$ and reduce as necessary. $y = \dfrac{1}{4}$ Substitute $\dfrac{1}{4}$ for $y$ in the top equation. $2x-8( \dfrac{1}{4}) = -8$ $2x-2 = -8$ $2x = -6$ $x = -3$ The solution is $\enspace x = -3, \enspace y = \dfrac{1}{4}$.